please help

Angle BCD = 80°, but don't ask me how I've done it.  

A regular heptagon and a regular 18-gon share a common edge AB, as shown.

Find angle BCD, in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\angle 1} &=& \dfrac{(7-2)180^\circ}{7}\mathbf{=\dfrac{5*180^\circ}{7}} \\ \hline 2*\angle 2 + \angle 1 &=& 180^\circ \\ 2*\angle 2 &=& 180^\circ - \angle 1 \\ 2*\angle 2 &=& 180^\circ - \dfrac{5*180^\circ}{7} \\ 2*\angle 2 &=& 180^\circ \left(1 - \dfrac{5}{7} \right) \\ 2*\angle 2 &=& 180^\circ \left( \dfrac{2}{7} \right) \\ \mathbf{\angle 2} &=& \mathbf{\dfrac{180^\circ}{7}} \\ \hline \mathbf{\angle 3} &=& \dfrac{(18-2)180^\circ}{18} =\dfrac{16*180^\circ}{18} \mathbf{=160^\circ } \\ \hline \angle 4 &=& 180^\circ - \angle 3 \\ \angle 4 &=& 180^\circ - 160^\circ \\ \mathbf{\angle 4} &=& \mathbf{20^\circ} \\ \hline \mathbf{\angle 5} &=& 180^\circ - \angle 1 \mathbf{= 180^\circ \left( \dfrac{2}{7} \right)} \\ \hline 2*\angle 6 + \angle 4 + \angle 5 &=& 180^\circ \\ 2*\angle 6 + 20^\circ + 180^\circ \left( \dfrac{2}{7} \right) &=& 180^\circ \\ 2*\angle 6 + 180^\circ \left( \dfrac{2}{7} \right) &=& 160^\circ \\ 2*\angle 6 &=& 160^\circ - 180^\circ \left( \dfrac{2}{7} \right) \\ \mathbf{\angle 6} &=& \mathbf{80^\circ - \dfrac{180^\circ}{7}} \\ \hline \angle BCD &=& \angle 2 + \angle 6 \\ \angle BCD &=& \dfrac{180^\circ}{7}+ 80^\circ - \dfrac{180^\circ}{7} \\ \mathbf{\angle BCD} &=& \mathbf{80^\circ} \\ \hline \end{array}\)